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This is Exercise 2.1 from Kirillov's Lie theory book.

Let $G$ be a Lie group and $H$ a closed Lie subgroup.

  1. Show that the closure $\overline{H}$ of $H$ in $G$ is a subgroup of $G$.
  1. Show that each coset $Hx$, $x\in\overline H$, is open and dense in $\overline H$.
  2. Show that $\overline H = H$.

Kirillov defines a closed Lie subgroup as

A closed Lie subgroup $H$ of a Lie group $G$ is a subgroup which is also an embedded submanifold.

I can show (1), the dense part of (2), and (3) assuming openness from (2). But how do I show that each $Hx$ is open in $\overline H$?

Or Shahar
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    Could someone explain how (3) follows from (2)? – Sha Vuklia Jan 16 '21 at 21:23
  • I also have this question for (3), I can prove (1) and (2) though. – Ruairi Feb 08 '21 at 21:01
  • @ShaVuklia Note that $H$ is open in $\overline{H}$, hence for any $x\in \overline{H}$ the intersection $H\cap Hx$ cannot be empty since $Hx$ is dense in $\overline{H}$. Some $gx\in H$ thus gives $x\in H$. – Shana Dec 07 '21 at 13:57

2 Answers2

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A small correction: part (1) of the problem asks to show that $\overline{H}$ is a subgroup of $G$.

Lemma: If $N$ is an (embedded) submanifold of $M$, then $N$ is locally closed. That is, every $x\in N$ has a neighborhood $V\subset M$ such that $N$ is closed in $V$ (i.e., $N\cap V=\overline{N}\cap V$).

By Lemma, every $h\in H$ has a neighborhood $V\subset G$ so that $H\cap V=\overline{H}\cap V$. By part (1), the diffeomorphism given by right multiplication by $x$ sends this set to $Hx\cap Vx=\overline{H}\cap Vx$, which is contained in $Hx$ and is open in $\overline{H}$; this shows that $Hx$ is open.

As for the Lemma, I learn the following proof from a problem set solution by Paul Apisa:

Proof of Lemma: Fix $x\in N$, and let $f:U\to\mathbb{R}^m$ be a chart of $x$ in $M$. Then $f|N:U\cap N\to\mathbb{R}^n$ is a chart of $x$ in $N$; set $h=f\circ(f|N)^{-1}$. Let $B$ be an open ball in $\mathbb{R}^n$ centered at $f(x)$, then $h(B)$ is open in $f(N)$, so there exists $U\subset M$ open so that $U\cap f(N)=h(B)$. Now $\overline{B}$ is compact, so $h(\overline{B})$ is compact and hence closed ($M$ is Hausdorff). Then $$ h(B) \subseteq U \cap h(\overline{B}) \subseteq U \cap f(N) = h(B), $$ so indeed $U\cap h(\overline{B})=h(B)=U\cap f(N)$. Therefore, $f^{-1}(U)\subset M$ is a neighborhood of $x$ such that $f^{-1}(U)\cap N=f^{-1}(U)\cap f^{-1}(h(\overline{B}))$ is closed in $f^{-1}(U)$, as desired.

Ruian Chen
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It suffices to show that $H\subseteq \overline{H}$ is open. For $x\in H,$ let $(\phi,U)$ be a slice chart for $H$ containing $x$, so that $\phi(U\cap H)=\phi(U)\cap \mathbb{R}^k\subseteq \mathbb{R}^n$ where $H$ is $k$-dimensional and $G$ is $n$-dimensional. This is possible because $H$ is embedded. Then none of the points in $U\backslash H$ are in the closure of $H$. It follows that $U\cap \overline{H}=U\cap H$. Therefore $U\cap H$ is an open neighborhood of $x$ in $\overline{H}$, relative to the subspace topology on $\overline{H}.$ Because $x$ was arbitrary, the conclusion follows.

subrosar
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